publicclassSolution{publicListNodeReverseList(ListNode head){if(head==null)returnnull;//head为当前节点,如果当前节点为空的话,那就什么也不做,直接返回null;ListNode pre=null;ListNode nextnode=null;while(head!=null){nextnode=head.next;head.next=pre;pre=head;head=nextnode;}returnpre;}} Reverse Lin...
* }*/classSolution {publicListNode reverseList(ListNode head) { ListNode first=head; ListNode reverseHead=null;//建立一个新的节点用来存放结果while(first !=null) {//遍历输入链表,开始处理每一个节点ListNode second = first.next;//先处理第一个节点first,所以需要一个指针来存储first的后继first.next ...
publicListNodereverseList5(ListNode head){if(head ==null|| head.next ==null) {returnhead; } ArrayList<Integer> list =newArrayList<Integer>();while(head !=null) { list.add(head.val); head = head.next; } ListNode node =newListNode(list.get(list.size()-1)); ListNode last = node;for...
Reverse Linked List II Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1->2->3->4->5->NULL, m = 2 and n = 4, return1->4->3->2->5->NULL. Note: Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of ...
Reverse a singly linked list. click to show more hints. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 既然问了能否iteratively or recursively, 那就both把. iterative 解法: 总结就是得到下一个节点,更改当前节点指向,将指针往下移动,直到过完整个linked...
Reverse a singly linked list. Example: Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL 1. 2. Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both? 反转链表,链表算法里必备技能点,基础的重中之重。很多有关链表反转、翻转的问...
def reverseList(self, head): """ :type head: ListNode :rtype: ListNode """ dummy = ListNode(None) while head: # 终止条件是head=Null nextnode = head.next # nextnode是head后面的节点 head.next = dummy # dummy.next是Null,所以这样head.next就成为了Null ...
public ListNode reverseList(ListNode head) { if (head == null || head.next == null){ return head; } ListNode p = reverseList(head.next); //重点 head.next.next = head; head.next = null; return p; } 总结 三种解法中其实利用栈来反转是最简单的,其次是迭代,最难的是递归,递归可以想成...
addFirst(list.get(i)); } return reversedList; } 效果有没有明显提升? problem avgt 403.000 us/op linked-list avgt 541 us/op 结果好多了! 第二种方法,保留ArrayList结构,在末尾插入元素(这样操作的时间复杂度是常数)。 必须对列表倒序迭代,如下所示: public List<Integer> reverseList(List<Integer...
问如何在java中反转链接列表，同时保持原来的顺序？EN说到json,相信没有人会陌生,我们天天都在用。那么...